分數規劃

 

poj2976

裸。。直接二分

#include<stdio.h>
#include<iostream>
#include<queue>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<math.h>
using namespace std;
int k,n;
double A[1010],B[1010],C[1010];
double f(double x){
    int i;
    double ans=0;
    for(i=1;i<=n;i++) C[i]=B[i]*x-A[i];
    sort(C+1,C+1+n);
    for(i=1;i<=k;i++) ans+=C[i];
    return ans;
}
int main(){
    int i;
    while(scanf("%d%d",&n,&k)!=EOF){
        if(n+k==0) break;
        k=n-k;
        for(i=1;i<=n;i++) scanf("%lf",&A[i]);
        for(i=1;i<=n;i++) scanf("%lf",&B[i]);

        double l=0,r=1e18,mid;
        while(l+1e-5<r){
             mid=(l+r)/2;
             if(f(mid)<0) l=mid;
             else r=mid;
        }
        mid*=100;
        long long ans=mid+0.5;
        printf("%lld\n",ans);
    }
    return 0;
}

 

bzoj4819 新生舞會

題面很裸的分數規劃。。直接二分，問題就成了求每次二分的最大（小）匹配權值，所以直接km。。

#include<stdio.h>
#include<iostream>
#include<queue>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<math.h>
using namespace std;
const int N=500;
double inf=1e18;
double e=1e-6;
struct KM{
    double hl[N],hr[N],slk[N],a[N][N];
    int fl[N],fr[N],vl[N],vr[N],pre[N],q[N],l,r,n;
    int check(int i){
        vl[i]=1;
        if(fl[i]){
            q[++r]=fl[i];
            vr[fl[i]]=1;
            return 1;
        }
        for(;i;swap(i,fr[pre[i]])) fl[i]=pre[i];
        return 0;
    }
    void bfs(int s){
        int i,j;
        double d;
        for(i=1;i<=n;i++){
            slk[i]=inf;
            vl[i]=vr[i]=0;
        }
        l=r=0;
        q[r]=s;
        vr[s]=1;
        while(1){
            while(l<=r){
                j=q[l++];
                for(i=1;i<=n;i++){
                    d=hl[i]+hr[j]-a[i][j];
                    if(vl[i]==0&&slk[i]>=d){
                        pre[i]=j;
                        if(fabs(d)>e) slk[i]=d;
                        else if(!check(i)) return;
                    }
                }
            }
            d=inf;
            for(i=1;i<=n;i++)if(!vl[i]&&d>slk[i]) d=slk[i];
            for(i=1;i<=n;i++){
                if(vl[i]) hl[i]+=d;
                else slk[i]-=d;
                if(vr[i]) hr[i]-=d;
            }
            for(i=1;i<=n;i++) if(!vl[i]&&slk[i]==0&&!check(i)) return;
        }
    }
    double ask(){
        int i,j;
        for(i=1;i<=n;i++){
            fl[i]=fr[i]=hr[i]=0;
            hl[i]=-inf;
        }
        for(i=1;i<=n;i++) for(j=1;j<=n;j++) hl[i]=max(a[i][j],hl[i]);
        for(j=1;j<=n;j++) bfs(j);
        double ans=0;
        for(i=1;i<=n;i++) ans+=a[i][fl[i]];
        return ans;
    }
}km;
 
 
int A[110][110],B[110][110];
double f(double x,int n){
    int i,j;
    km.n=n;
    for(i=1;i<=n;i++) for(j=1;j<=n;j++) km.a[i][j]=-B[i][j]*x+A[i][j];
    return -km.ask();
}
 
 
int main(){
    int i,j,n;
    scanf("%d",&n);
    for(i=1;i<=n;i++) for(j=1;j<=n;j++) scanf("%d",&A[i][j]);
    for(i=1;i<=n;i++) for(j=1;j<=n;j++) scanf("%d",&B[i][j]);
    double l=0,r=1e9,mid;
    while(l+1e-10<r){
        mid=(l+r)/2;
        if(f(mid,n)<0) l=mid;
        else r=mid;
    }
    printf("%.6f\n",mid);
    return 0;
}